A friend claims, “I only played that lottery \(n=25\) times, and I’ve won \(\sum x = 35\) dollars.” You wish to judge whether your friend is making an outlandish claim. To do this, you calculate a \(z\) score.
\[z = \frac{\sum x - n\mu}{\sigma\sqrt{n}}\]
A typical \(z\) score is between \(-2\) and \(2\) (this happens 95% of the time). A \(z\) score less than \(-3\) or more than \(3\) is quite unlikely (less than 0.3% of the time).
Is your friend’s claim typical or outlandish? To answer this, calculate a \(z\) score accurate to the hundredths place.
\[z=\,?\]
Solution
Use the link to determine the population mean \(\mu=-0.2\) and population standard deviation \(\sigma = 4.069398\).
A friend claims, “I only played that lottery \(n=25\) times, and I’ve won \(\sum x = 68\) dollars.” You wish to judge whether your friend is making an outlandish claim. To do this, you calculate a \(z\) score.
\[z = \frac{\sum x - n\mu}{\sigma\sqrt{n}}\]
A typical \(z\) score is between \(-2\) and \(2\) (this happens 95% of the time). A \(z\) score less than \(-3\) or more than \(3\) is quite unlikely (less than 0.3% of the time).
Is your friend’s claim typical or outlandish? To answer this, calculate a \(z\) score accurate to the hundredths place.
\[z=\,?\]
Solution
Use the link to determine the population mean \(\mu=-0.7\) and population standard deviation \(\sigma = 5.1778374\).
A friend claims, “I only played that lottery \(n=25\) times, and I’ve won \(\sum x = 21\) dollars.” You wish to judge whether your friend is making an outlandish claim. To do this, you calculate a \(z\) score.
\[z = \frac{\sum x - n\mu}{\sigma\sqrt{n}}\]
A typical \(z\) score is between \(-2\) and \(2\) (this happens 95% of the time). A \(z\) score less than \(-3\) or more than \(3\) is quite unlikely (less than 0.3% of the time).
Is your friend’s claim typical or outlandish? To answer this, calculate a \(z\) score accurate to the hundredths place.
\[z=\,?\]
Solution
Use the link to determine the population mean \(\mu=-0.4\) and population standard deviation \(\sigma = 4.9234135\).
A friend claims, “I only played that lottery \(n=25\) times, and I’ve won \(\sum x = 24\) dollars.” You wish to judge whether your friend is making an outlandish claim. To do this, you calculate a \(z\) score.
\[z = \frac{\sum x - n\mu}{\sigma\sqrt{n}}\]
A typical \(z\) score is between \(-2\) and \(2\) (this happens 95% of the time). A \(z\) score less than \(-3\) or more than \(3\) is quite unlikely (less than 0.3% of the time).
Is your friend’s claim typical or outlandish? To answer this, calculate a \(z\) score accurate to the hundredths place.
\[z=\,?\]
Solution
Use the link to determine the population mean \(\mu=-0.5\) and population standard deviation \(\sigma = 5.3338541\).
A friend claims, “I only played that lottery \(n=25\) times, and I’ve won \(\sum x = 113\) dollars.” You wish to judge whether your friend is making an outlandish claim. To do this, you calculate a \(z\) score.
\[z = \frac{\sum x - n\mu}{\sigma\sqrt{n}}\]
A typical \(z\) score is between \(-2\) and \(2\) (this happens 95% of the time). A \(z\) score less than \(-3\) or more than \(3\) is quite unlikely (less than 0.3% of the time).
Is your friend’s claim typical or outlandish? To answer this, calculate a \(z\) score accurate to the hundredths place.
\[z=\,?\]
Solution
Use the link to determine the population mean \(\mu=-0.1\) and population standard deviation \(\sigma = 6.5795137\).
A friend claims, “I only played that lottery \(n=25\) times, and I’ve won \(\sum x = 29\) dollars.” You wish to judge whether your friend is making an outlandish claim. To do this, you calculate a \(z\) score.
\[z = \frac{\sum x - n\mu}{\sigma\sqrt{n}}\]
A typical \(z\) score is between \(-2\) and \(2\) (this happens 95% of the time). A \(z\) score less than \(-3\) or more than \(3\) is quite unlikely (less than 0.3% of the time).
Is your friend’s claim typical or outlandish? To answer this, calculate a \(z\) score accurate to the hundredths place.
\[z=\,?\]
Solution
Use the link to determine the population mean \(\mu=-0.7\) and population standard deviation \(\sigma = 3.0675723\).
A friend claims, “I only played that lottery \(n=25\) times, and I’ve won \(\sum x = 24\) dollars.” You wish to judge whether your friend is making an outlandish claim. To do this, you calculate a \(z\) score.
\[z = \frac{\sum x - n\mu}{\sigma\sqrt{n}}\]
A typical \(z\) score is between \(-2\) and \(2\) (this happens 95% of the time). A \(z\) score less than \(-3\) or more than \(3\) is quite unlikely (less than 0.3% of the time).
Is your friend’s claim typical or outlandish? To answer this, calculate a \(z\) score accurate to the hundredths place.
\[z=\,?\]
Solution
Use the link to determine the population mean \(\mu=-0.2\) and population standard deviation \(\sigma = 4.8124838\).
A friend claims, “I only played that lottery \(n=25\) times, and I’ve won \(\sum x = 75\) dollars.” You wish to judge whether your friend is making an outlandish claim. To do this, you calculate a \(z\) score.
\[z = \frac{\sum x - n\mu}{\sigma\sqrt{n}}\]
A typical \(z\) score is between \(-2\) and \(2\) (this happens 95% of the time). A \(z\) score less than \(-3\) or more than \(3\) is quite unlikely (less than 0.3% of the time).
Is your friend’s claim typical or outlandish? To answer this, calculate a \(z\) score accurate to the hundredths place.
\[z=\,?\]
Solution
Use the link to determine the population mean \(\mu=-0.1\) and population standard deviation \(\sigma = 4.5486262\).
A friend claims, “I only played that lottery \(n=25\) times, and I’ve won \(\sum x = 66\) dollars.” You wish to judge whether your friend is making an outlandish claim. To do this, you calculate a \(z\) score.
\[z = \frac{\sum x - n\mu}{\sigma\sqrt{n}}\]
A typical \(z\) score is between \(-2\) and \(2\) (this happens 95% of the time). A \(z\) score less than \(-3\) or more than \(3\) is quite unlikely (less than 0.3% of the time).
Is your friend’s claim typical or outlandish? To answer this, calculate a \(z\) score accurate to the hundredths place.
\[z=\,?\]
Solution
Use the link to determine the population mean \(\mu=-0.8\) and population standard deviation \(\sigma = 5.3814496\).
A friend claims, “I only played that lottery \(n=25\) times, and I’ve won \(\sum x = 36\) dollars.” You wish to judge whether your friend is making an outlandish claim. To do this, you calculate a \(z\) score.
\[z = \frac{\sum x - n\mu}{\sigma\sqrt{n}}\]
A typical \(z\) score is between \(-2\) and \(2\) (this happens 95% of the time). A \(z\) score less than \(-3\) or more than \(3\) is quite unlikely (less than 0.3% of the time).
Is your friend’s claim typical or outlandish? To answer this, calculate a \(z\) score accurate to the hundredths place.
\[z=\,?\]
Solution
Use the link to determine the population mean \(\mu=-0.4\) and population standard deviation \(\sigma = 5.2763624\).
The numbers that satisfy \(Z<-0.62\) are on the left side of a number line (toward \(-\infty\)). The probability equals a left area under the density curve.
By using the z-table, we find the appropriate probability.
\[P(Z<-0.62)=0.2676\]
It might help to visualize with a spinner:
Using a spreadsheet:
=NORM.DIST(-0.62,0,1,TRUE)
Using R:
pnorm(-0.62)
Question
Determine the probability that the standard normal variable is less than -1.05. In other words, evaluate \(P(Z < -1.05)\).
The numbers that satisfy \(Z<-1.05\) are on the left side of a number line (toward \(-\infty\)). The probability equals a left area under the density curve.
By using the z-table, we find the appropriate probability.
\[P(Z<-1.05)=0.1469\]
It might help to visualize with a spinner:
Using a spreadsheet:
=NORM.DIST(-1.05,0,1,TRUE)
Using R:
pnorm(-1.05)
Question
Determine the probability that the standard normal variable is less than 0.51. In other words, evaluate \(P(Z < 0.51)\).
The numbers that satisfy \(Z<0.51\) are on the left side of a number line (toward \(-\infty\)). The probability equals a left area under the density curve.
By using the z-table, we find the appropriate probability.
\[P(Z<0.51)=0.695\]
It might help to visualize with a spinner:
Using a spreadsheet:
=NORM.DIST(0.51,0,1,TRUE)
Using R:
pnorm(0.51)
Question
Determine the probability that the standard normal variable is less than 0.36. In other words, evaluate \(P(Z < 0.36)\).
The numbers that satisfy \(Z<0.36\) are on the left side of a number line (toward \(-\infty\)). The probability equals a left area under the density curve.
By using the z-table, we find the appropriate probability.
\[P(Z<0.36)=0.6406\]
It might help to visualize with a spinner:
Using a spreadsheet:
=NORM.DIST(0.36,0,1,TRUE)
Using R:
pnorm(0.36)
Question
Determine the probability that the standard normal variable is less than -0.54. In other words, evaluate \(P(Z < -0.54)\).
The numbers that satisfy \(Z<-0.54\) are on the left side of a number line (toward \(-\infty\)). The probability equals a left area under the density curve.
By using the z-table, we find the appropriate probability.
\[P(Z<-0.54)=0.2946\]
It might help to visualize with a spinner:
Using a spreadsheet:
=NORM.DIST(-0.54,0,1,TRUE)
Using R:
pnorm(-0.54)
Question
Determine the probability that the standard normal variable is less than 0.58. In other words, evaluate \(P(Z < 0.58)\).
The numbers that satisfy \(Z<0.58\) are on the left side of a number line (toward \(-\infty\)). The probability equals a left area under the density curve.
By using the z-table, we find the appropriate probability.
\[P(Z<0.58)=0.719\]
It might help to visualize with a spinner:
Using a spreadsheet:
=NORM.DIST(0.58,0,1,TRUE)
Using R:
pnorm(0.58)
Question
Determine the probability that the standard normal variable is less than -0.59. In other words, evaluate \(P(Z < -0.59)\).
The numbers that satisfy \(Z<-0.59\) are on the left side of a number line (toward \(-\infty\)). The probability equals a left area under the density curve.
By using the z-table, we find the appropriate probability.
\[P(Z<-0.59)=0.2776\]
It might help to visualize with a spinner:
Using a spreadsheet:
=NORM.DIST(-0.59,0,1,TRUE)
Using R:
pnorm(-0.59)
Question
Determine the probability that the standard normal variable is less than 1.76. In other words, evaluate \(P(Z < 1.76)\).
The numbers that satisfy \(Z<1.76\) are on the left side of a number line (toward \(-\infty\)). The probability equals a left area under the density curve.
By using the z-table, we find the appropriate probability.
\[P(Z<1.76)=0.9608\]
It might help to visualize with a spinner:
Using a spreadsheet:
=NORM.DIST(1.76,0,1,TRUE)
Using R:
pnorm(1.76)
Question
Determine the probability that the standard normal variable is less than 0.99. In other words, evaluate \(P(Z < 0.99)\).
The numbers that satisfy \(Z<0.99\) are on the left side of a number line (toward \(-\infty\)). The probability equals a left area under the density curve.
By using the z-table, we find the appropriate probability.
\[P(Z<0.99)=0.8389\]
It might help to visualize with a spinner:
Using a spreadsheet:
=NORM.DIST(0.99,0,1,TRUE)
Using R:
pnorm(0.99)
Question
Determine the probability that the standard normal variable is less than 0.77. In other words, evaluate \(P(Z < 0.77)\).
The numbers that satisfy \(Z<0.77\) are on the left side of a number line (toward \(-\infty\)). The probability equals a left area under the density curve.
By using the z-table, we find the appropriate probability.
\[P(Z<0.77)=0.7794\]
It might help to visualize with a spinner:
Using a spreadsheet:
=NORM.DIST(0.77,0,1,TRUE)
Using R:
pnorm(0.77)
Question
Determine the probability that the standard normal variable is more than -1.36. In other words, evaluate \(P(Z > -1.36)\).
First, you need to identify that we are looking for a central area. This is because \(z\)-scores near 0 satisfy \(|Z|<0.68\), while \(z\)-scores far from 0 (either positive or negative) do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provide left areas, so three methods are shown to get the central area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-0.68) = 0.2483\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>0.68) = 0.2483 \]
The three areas add to 1.
\[\begin{aligned}
P\left(|Z|<0.68\right) &= 1-P(Z<-0.68)-P(Z>0.68) \\
&= 1 - 0.2483 - 0.2483 \\
&= 0.5034
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|<z\right) = 1-2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=0.68\)),
\[\begin{aligned}
P\left(|Z|<0.68\right) &= 1-2\cdot P(Z<-0.68) \\
&= 1-2\cdot0.2483 \\
&= 0.5034
\end{aligned}\]
This method is shown graphically:
Method 2: You can also find half of the central area and then double it.
\[\begin{aligned}
P\left(|Z|<z\right) &= 2\cdot\left[ P(Z<z)-0.5\right] \\
&= 2\cdot\left[ 0.7517-0.5\right] \\
&= 2\cdot\left[ 0.2517\right] \\
&= 0.5034
\end{aligned}\]
This method is shown graphically:
Method 3: You can also calculate the central area with a difference of two left areas.
First, you need to identify that we are looking for a central area. This is because \(z\)-scores near 0 satisfy \(|Z|<1.04\), while \(z\)-scores far from 0 (either positive or negative) do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provide left areas, so three methods are shown to get the central area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-1.04) = 0.1492\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>1.04) = 0.1492 \]
The three areas add to 1.
\[\begin{aligned}
P\left(|Z|<1.04\right) &= 1-P(Z<-1.04)-P(Z>1.04) \\
&= 1 - 0.1492 - 0.1492 \\
&= 0.7016
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|<z\right) = 1-2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=1.04\)),
\[\begin{aligned}
P\left(|Z|<1.04\right) &= 1-2\cdot P(Z<-1.04) \\
&= 1-2\cdot0.1492 \\
&= 0.7016
\end{aligned}\]
This method is shown graphically:
Method 2: You can also find half of the central area and then double it.
\[\begin{aligned}
P\left(|Z|<z\right) &= 2\cdot\left[ P(Z<z)-0.5\right] \\
&= 2\cdot\left[ 0.8508-0.5\right] \\
&= 2\cdot\left[ 0.3508\right] \\
&= 0.7016
\end{aligned}\]
This method is shown graphically:
Method 3: You can also calculate the central area with a difference of two left areas.
First, you need to identify that we are looking for a central area. This is because \(z\)-scores near 0 satisfy \(|Z|<0.69\), while \(z\)-scores far from 0 (either positive or negative) do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provide left areas, so three methods are shown to get the central area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-0.69) = 0.2451\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>0.69) = 0.2451 \]
The three areas add to 1.
\[\begin{aligned}
P\left(|Z|<0.69\right) &= 1-P(Z<-0.69)-P(Z>0.69) \\
&= 1 - 0.2451 - 0.2451 \\
&= 0.5098
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|<z\right) = 1-2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=0.69\)),
\[\begin{aligned}
P\left(|Z|<0.69\right) &= 1-2\cdot P(Z<-0.69) \\
&= 1-2\cdot0.2451 \\
&= 0.5098
\end{aligned}\]
This method is shown graphically:
Method 2: You can also find half of the central area and then double it.
\[\begin{aligned}
P\left(|Z|<z\right) &= 2\cdot\left[ P(Z<z)-0.5\right] \\
&= 2\cdot\left[ 0.7549-0.5\right] \\
&= 2\cdot\left[ 0.2549\right] \\
&= 0.5098
\end{aligned}\]
This method is shown graphically:
Method 3: You can also calculate the central area with a difference of two left areas.
First, you need to identify that we are looking for a central area. This is because \(z\)-scores near 0 satisfy \(|Z|<0.96\), while \(z\)-scores far from 0 (either positive or negative) do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provide left areas, so three methods are shown to get the central area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-0.96) = 0.1685\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>0.96) = 0.1685 \]
The three areas add to 1.
\[\begin{aligned}
P\left(|Z|<0.96\right) &= 1-P(Z<-0.96)-P(Z>0.96) \\
&= 1 - 0.1685 - 0.1685 \\
&= 0.663
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|<z\right) = 1-2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=0.96\)),
\[\begin{aligned}
P\left(|Z|<0.96\right) &= 1-2\cdot P(Z<-0.96) \\
&= 1-2\cdot0.1685 \\
&= 0.663
\end{aligned}\]
This method is shown graphically:
Method 2: You can also find half of the central area and then double it.
\[\begin{aligned}
P\left(|Z|<z\right) &= 2\cdot\left[ P(Z<z)-0.5\right] \\
&= 2\cdot\left[ 0.8315-0.5\right] \\
&= 2\cdot\left[ 0.3315\right] \\
&= 0.663
\end{aligned}\]
This method is shown graphically:
Method 3: You can also calculate the central area with a difference of two left areas.
First, you need to identify that we are looking for a central area. This is because \(z\)-scores near 0 satisfy \(|Z|<0.57\), while \(z\)-scores far from 0 (either positive or negative) do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provide left areas, so three methods are shown to get the central area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-0.57) = 0.2843\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>0.57) = 0.2843 \]
The three areas add to 1.
\[\begin{aligned}
P\left(|Z|<0.57\right) &= 1-P(Z<-0.57)-P(Z>0.57) \\
&= 1 - 0.2843 - 0.2843 \\
&= 0.4314
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|<z\right) = 1-2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=0.57\)),
\[\begin{aligned}
P\left(|Z|<0.57\right) &= 1-2\cdot P(Z<-0.57) \\
&= 1-2\cdot0.2843 \\
&= 0.4314
\end{aligned}\]
This method is shown graphically:
Method 2: You can also find half of the central area and then double it.
\[\begin{aligned}
P\left(|Z|<z\right) &= 2\cdot\left[ P(Z<z)-0.5\right] \\
&= 2\cdot\left[ 0.7157-0.5\right] \\
&= 2\cdot\left[ 0.2157\right] \\
&= 0.4314
\end{aligned}\]
This method is shown graphically:
Method 3: You can also calculate the central area with a difference of two left areas.
First, you need to identify that we are looking for a central area. This is because \(z\)-scores near 0 satisfy \(|Z|<1.29\), while \(z\)-scores far from 0 (either positive or negative) do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provide left areas, so three methods are shown to get the central area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-1.29) = 0.0985\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>1.29) = 0.0985 \]
The three areas add to 1.
\[\begin{aligned}
P\left(|Z|<1.29\right) &= 1-P(Z<-1.29)-P(Z>1.29) \\
&= 1 - 0.0985 - 0.0985 \\
&= 0.803
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|<z\right) = 1-2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=1.29\)),
\[\begin{aligned}
P\left(|Z|<1.29\right) &= 1-2\cdot P(Z<-1.29) \\
&= 1-2\cdot0.0985 \\
&= 0.803
\end{aligned}\]
This method is shown graphically:
Method 2: You can also find half of the central area and then double it.
\[\begin{aligned}
P\left(|Z|<z\right) &= 2\cdot\left[ P(Z<z)-0.5\right] \\
&= 2\cdot\left[ 0.9015-0.5\right] \\
&= 2\cdot\left[ 0.4015\right] \\
&= 0.803
\end{aligned}\]
This method is shown graphically:
Method 3: You can also calculate the central area with a difference of two left areas.
First, you need to identify that we are looking for a central area. This is because \(z\)-scores near 0 satisfy \(|Z|<1.1\), while \(z\)-scores far from 0 (either positive or negative) do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provide left areas, so three methods are shown to get the central area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-1.1) = 0.1357\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>1.1) = 0.1357 \]
The three areas add to 1.
\[\begin{aligned}
P\left(|Z|<1.1\right) &= 1-P(Z<-1.1)-P(Z>1.1) \\
&= 1 - 0.1357 - 0.1357 \\
&= 0.7286
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|<z\right) = 1-2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=1.1\)),
\[\begin{aligned}
P\left(|Z|<1.1\right) &= 1-2\cdot P(Z<-1.1) \\
&= 1-2\cdot0.1357 \\
&= 0.7286
\end{aligned}\]
This method is shown graphically:
Method 2: You can also find half of the central area and then double it.
\[\begin{aligned}
P\left(|Z|<z\right) &= 2\cdot\left[ P(Z<z)-0.5\right] \\
&= 2\cdot\left[ 0.8643-0.5\right] \\
&= 2\cdot\left[ 0.3643\right] \\
&= 0.7286
\end{aligned}\]
This method is shown graphically:
Method 3: You can also calculate the central area with a difference of two left areas.
First, you need to identify that we are looking for a central area. This is because \(z\)-scores near 0 satisfy \(|Z|<1.32\), while \(z\)-scores far from 0 (either positive or negative) do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provide left areas, so three methods are shown to get the central area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-1.32) = 0.0934\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>1.32) = 0.0934 \]
The three areas add to 1.
\[\begin{aligned}
P\left(|Z|<1.32\right) &= 1-P(Z<-1.32)-P(Z>1.32) \\
&= 1 - 0.0934 - 0.0934 \\
&= 0.8132
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|<z\right) = 1-2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=1.32\)),
\[\begin{aligned}
P\left(|Z|<1.32\right) &= 1-2\cdot P(Z<-1.32) \\
&= 1-2\cdot0.0934 \\
&= 0.8132
\end{aligned}\]
This method is shown graphically:
Method 2: You can also find half of the central area and then double it.
\[\begin{aligned}
P\left(|Z|<z\right) &= 2\cdot\left[ P(Z<z)-0.5\right] \\
&= 2\cdot\left[ 0.9066-0.5\right] \\
&= 2\cdot\left[ 0.4066\right] \\
&= 0.8132
\end{aligned}\]
This method is shown graphically:
Method 3: You can also calculate the central area with a difference of two left areas.
First, you need to identify that we are looking for a central area. This is because \(z\)-scores near 0 satisfy \(|Z|<0.62\), while \(z\)-scores far from 0 (either positive or negative) do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provide left areas, so three methods are shown to get the central area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-0.62) = 0.2676\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>0.62) = 0.2676 \]
The three areas add to 1.
\[\begin{aligned}
P\left(|Z|<0.62\right) &= 1-P(Z<-0.62)-P(Z>0.62) \\
&= 1 - 0.2676 - 0.2676 \\
&= 0.4648
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|<z\right) = 1-2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=0.62\)),
\[\begin{aligned}
P\left(|Z|<0.62\right) &= 1-2\cdot P(Z<-0.62) \\
&= 1-2\cdot0.2676 \\
&= 0.4648
\end{aligned}\]
This method is shown graphically:
Method 2: You can also find half of the central area and then double it.
\[\begin{aligned}
P\left(|Z|<z\right) &= 2\cdot\left[ P(Z<z)-0.5\right] \\
&= 2\cdot\left[ 0.7324-0.5\right] \\
&= 2\cdot\left[ 0.2324\right] \\
&= 0.4648
\end{aligned}\]
This method is shown graphically:
Method 3: You can also calculate the central area with a difference of two left areas.
First, you need to identify that we are looking for a central area. This is because \(z\)-scores near 0 satisfy \(|Z|<0.44\), while \(z\)-scores far from 0 (either positive or negative) do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provide left areas, so three methods are shown to get the central area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-0.44) = 0.33\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>0.44) = 0.33 \]
The three areas add to 1.
\[\begin{aligned}
P\left(|Z|<0.44\right) &= 1-P(Z<-0.44)-P(Z>0.44) \\
&= 1 - 0.33 - 0.33 \\
&= 0.34
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|<z\right) = 1-2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=0.44\)),
\[\begin{aligned}
P\left(|Z|<0.44\right) &= 1-2\cdot P(Z<-0.44) \\
&= 1-2\cdot0.33 \\
&= 0.34
\end{aligned}\]
This method is shown graphically:
Method 2: You can also find half of the central area and then double it.
\[\begin{aligned}
P\left(|Z|<z\right) &= 2\cdot\left[ P(Z<z)-0.5\right] \\
&= 2\cdot\left[ 0.67-0.5\right] \\
&= 2\cdot\left[ 0.17\right] \\
&= 0.34
\end{aligned}\]
This method is shown graphically:
Method 3: You can also calculate the central area with a difference of two left areas.
First, you need to identify that we are looking for a two-tail area (the sum of the left and right tails). This is because \(z\)-scores far from 0 satisfy \(|Z|>0.57\), while \(z\)-scores near 0 do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provides left areas, so three methods are shown to get the two-tail area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-0.57) = 0.2843\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>0.57) = 0.2843 \]
The two areas add to our desired two-tail area.
\[\begin{aligned}
P\left(|Z|>0.57\right) &= P(Z<-0.57)+P(Z>0.57) \\
&= 0.2843 + 0.2843 \\
&= 0.5686
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|>z\right) = 2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=0.57\)),
\[\begin{aligned}
P\left(|Z|<0.57\right) &= 2\cdot P(Z<-0.57) \\
&= 2\cdot0.2843 \\
&= 0.5686
\end{aligned}\]
This method is shown graphically:
Notice we need to add both tails.
Method 2: You can achieve the same result by using the following formula:
\[P\left(|Z|>z\right) = 2\cdot\left[1-P(Z<z) \right] \]
So,
\[\begin{aligned}
P\left(|Z|>0.57\right) &= 2\cdot\left[1-P(Z<0.57) \right] \\
&= 2\cdot\left[1-0.7157\right] \\
&= 2\cdot0.2843\\
&= 0.5686
\end{aligned}\]
It might be helpful to visualize with a spinner.
In a spreadsheet, you could use the NORM.DIST() function.
=2*NORM.DIST(-0.57,0,1,TRUE)
In R, you could use the pnorm function.
2*pnorm(-0.57)
Question
Determine the probability that the absolute standard normal variable is more than 0.85. In other words, evaluate \(P\left(|Z| > 0.85\right)\).
First, you need to identify that we are looking for a two-tail area (the sum of the left and right tails). This is because \(z\)-scores far from 0 satisfy \(|Z|>0.85\), while \(z\)-scores near 0 do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provides left areas, so three methods are shown to get the two-tail area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-0.85) = 0.1977\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>0.85) = 0.1977 \]
The two areas add to our desired two-tail area.
\[\begin{aligned}
P\left(|Z|>0.85\right) &= P(Z<-0.85)+P(Z>0.85) \\
&= 0.1977 + 0.1977 \\
&= 0.3954
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|>z\right) = 2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=0.85\)),
\[\begin{aligned}
P\left(|Z|<0.85\right) &= 2\cdot P(Z<-0.85) \\
&= 2\cdot0.1977 \\
&= 0.3954
\end{aligned}\]
This method is shown graphically:
Notice we need to add both tails.
Method 2: You can achieve the same result by using the following formula:
\[P\left(|Z|>z\right) = 2\cdot\left[1-P(Z<z) \right] \]
So,
\[\begin{aligned}
P\left(|Z|>0.85\right) &= 2\cdot\left[1-P(Z<0.85) \right] \\
&= 2\cdot\left[1-0.8023\right] \\
&= 2\cdot0.1977\\
&= 0.3954
\end{aligned}\]
It might be helpful to visualize with a spinner.
In a spreadsheet, you could use the NORM.DIST() function.
=2*NORM.DIST(-0.85,0,1,TRUE)
In R, you could use the pnorm function.
2*pnorm(-0.85)
Question
Determine the probability that the absolute standard normal variable is more than 1.12. In other words, evaluate \(P\left(|Z| > 1.12\right)\).
First, you need to identify that we are looking for a two-tail area (the sum of the left and right tails). This is because \(z\)-scores far from 0 satisfy \(|Z|>1.12\), while \(z\)-scores near 0 do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provides left areas, so three methods are shown to get the two-tail area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-1.12) = 0.1314\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>1.12) = 0.1314 \]
The two areas add to our desired two-tail area.
\[\begin{aligned}
P\left(|Z|>1.12\right) &= P(Z<-1.12)+P(Z>1.12) \\
&= 0.1314 + 0.1314 \\
&= 0.2628
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|>z\right) = 2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=1.12\)),
\[\begin{aligned}
P\left(|Z|<1.12\right) &= 2\cdot P(Z<-1.12) \\
&= 2\cdot0.1314 \\
&= 0.2628
\end{aligned}\]
This method is shown graphically:
Notice we need to add both tails.
Method 2: You can achieve the same result by using the following formula:
\[P\left(|Z|>z\right) = 2\cdot\left[1-P(Z<z) \right] \]
So,
\[\begin{aligned}
P\left(|Z|>1.12\right) &= 2\cdot\left[1-P(Z<1.12) \right] \\
&= 2\cdot\left[1-0.8686\right] \\
&= 2\cdot0.1314\\
&= 0.2628
\end{aligned}\]
It might be helpful to visualize with a spinner.
In a spreadsheet, you could use the NORM.DIST() function.
=2*NORM.DIST(-1.12,0,1,TRUE)
In R, you could use the pnorm function.
2*pnorm(-1.12)
Question
Determine the probability that the absolute standard normal variable is more than 1.75. In other words, evaluate \(P\left(|Z| > 1.75\right)\).
First, you need to identify that we are looking for a two-tail area (the sum of the left and right tails). This is because \(z\)-scores far from 0 satisfy \(|Z|>1.75\), while \(z\)-scores near 0 do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provides left areas, so three methods are shown to get the two-tail area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-1.75) = 0.0401\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>1.75) = 0.0401 \]
The two areas add to our desired two-tail area.
\[\begin{aligned}
P\left(|Z|>1.75\right) &= P(Z<-1.75)+P(Z>1.75) \\
&= 0.0401 + 0.0401 \\
&= 0.0802
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|>z\right) = 2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=1.75\)),
\[\begin{aligned}
P\left(|Z|<1.75\right) &= 2\cdot P(Z<-1.75) \\
&= 2\cdot0.0401 \\
&= 0.0802
\end{aligned}\]
This method is shown graphically:
Notice we need to add both tails.
Method 2: You can achieve the same result by using the following formula:
\[P\left(|Z|>z\right) = 2\cdot\left[1-P(Z<z) \right] \]
So,
\[\begin{aligned}
P\left(|Z|>1.75\right) &= 2\cdot\left[1-P(Z<1.75) \right] \\
&= 2\cdot\left[1-0.9599\right] \\
&= 2\cdot0.0401\\
&= 0.0802
\end{aligned}\]
It might be helpful to visualize with a spinner.
In a spreadsheet, you could use the NORM.DIST() function.
=2*NORM.DIST(-1.75,0,1,TRUE)
In R, you could use the pnorm function.
2*pnorm(-1.75)
Question
Determine the probability that the absolute standard normal variable is more than 1.7. In other words, evaluate \(P\left(|Z| > 1.7\right)\).
First, you need to identify that we are looking for a two-tail area (the sum of the left and right tails). This is because \(z\)-scores far from 0 satisfy \(|Z|>1.7\), while \(z\)-scores near 0 do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provides left areas, so three methods are shown to get the two-tail area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-1.7) = 0.0446\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>1.7) = 0.0446 \]
The two areas add to our desired two-tail area.
\[\begin{aligned}
P\left(|Z|>1.7\right) &= P(Z<-1.7)+P(Z>1.7) \\
&= 0.0446 + 0.0446 \\
&= 0.0892
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|>z\right) = 2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=1.7\)),
\[\begin{aligned}
P\left(|Z|<1.7\right) &= 2\cdot P(Z<-1.7) \\
&= 2\cdot0.0446 \\
&= 0.0892
\end{aligned}\]
This method is shown graphically:
Notice we need to add both tails.
Method 2: You can achieve the same result by using the following formula:
\[P\left(|Z|>z\right) = 2\cdot\left[1-P(Z<z) \right] \]
So,
\[\begin{aligned}
P\left(|Z|>1.7\right) &= 2\cdot\left[1-P(Z<1.7) \right] \\
&= 2\cdot\left[1-0.9554\right] \\
&= 2\cdot0.0446\\
&= 0.0892
\end{aligned}\]
It might be helpful to visualize with a spinner.
In a spreadsheet, you could use the NORM.DIST() function.
=2*NORM.DIST(-1.7,0,1,TRUE)
In R, you could use the pnorm function.
2*pnorm(-1.7)
Question
Determine the probability that the absolute standard normal variable is more than 0.99. In other words, evaluate \(P\left(|Z| > 0.99\right)\).
First, you need to identify that we are looking for a two-tail area (the sum of the left and right tails). This is because \(z\)-scores far from 0 satisfy \(|Z|>0.99\), while \(z\)-scores near 0 do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provides left areas, so three methods are shown to get the two-tail area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-0.99) = 0.1611\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>0.99) = 0.1611 \]
The two areas add to our desired two-tail area.
\[\begin{aligned}
P\left(|Z|>0.99\right) &= P(Z<-0.99)+P(Z>0.99) \\
&= 0.1611 + 0.1611 \\
&= 0.3222
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|>z\right) = 2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=0.99\)),
\[\begin{aligned}
P\left(|Z|<0.99\right) &= 2\cdot P(Z<-0.99) \\
&= 2\cdot0.1611 \\
&= 0.3222
\end{aligned}\]
This method is shown graphically:
Notice we need to add both tails.
Method 2: You can achieve the same result by using the following formula:
\[P\left(|Z|>z\right) = 2\cdot\left[1-P(Z<z) \right] \]
So,
\[\begin{aligned}
P\left(|Z|>0.99\right) &= 2\cdot\left[1-P(Z<0.99) \right] \\
&= 2\cdot\left[1-0.8389\right] \\
&= 2\cdot0.1611\\
&= 0.3222
\end{aligned}\]
It might be helpful to visualize with a spinner.
In a spreadsheet, you could use the NORM.DIST() function.
=2*NORM.DIST(-0.99,0,1,TRUE)
In R, you could use the pnorm function.
2*pnorm(-0.99)
Question
Determine the probability that the absolute standard normal variable is more than 0.83. In other words, evaluate \(P\left(|Z| > 0.83\right)\).
First, you need to identify that we are looking for a two-tail area (the sum of the left and right tails). This is because \(z\)-scores far from 0 satisfy \(|Z|>0.83\), while \(z\)-scores near 0 do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provides left areas, so three methods are shown to get the two-tail area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-0.83) = 0.2033\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>0.83) = 0.2033 \]
The two areas add to our desired two-tail area.
\[\begin{aligned}
P\left(|Z|>0.83\right) &= P(Z<-0.83)+P(Z>0.83) \\
&= 0.2033 + 0.2033 \\
&= 0.4066
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|>z\right) = 2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=0.83\)),
\[\begin{aligned}
P\left(|Z|<0.83\right) &= 2\cdot P(Z<-0.83) \\
&= 2\cdot0.2033 \\
&= 0.4066
\end{aligned}\]
This method is shown graphically:
Notice we need to add both tails.
Method 2: You can achieve the same result by using the following formula:
\[P\left(|Z|>z\right) = 2\cdot\left[1-P(Z<z) \right] \]
So,
\[\begin{aligned}
P\left(|Z|>0.83\right) &= 2\cdot\left[1-P(Z<0.83) \right] \\
&= 2\cdot\left[1-0.7967\right] \\
&= 2\cdot0.2033\\
&= 0.4066
\end{aligned}\]
It might be helpful to visualize with a spinner.
In a spreadsheet, you could use the NORM.DIST() function.
=2*NORM.DIST(-0.83,0,1,TRUE)
In R, you could use the pnorm function.
2*pnorm(-0.83)
Question
Determine the probability that the absolute standard normal variable is more than 1.8. In other words, evaluate \(P\left(|Z| > 1.8\right)\).
First, you need to identify that we are looking for a two-tail area (the sum of the left and right tails). This is because \(z\)-scores far from 0 satisfy \(|Z|>1.8\), while \(z\)-scores near 0 do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provides left areas, so three methods are shown to get the two-tail area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-1.8) = 0.0359\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>1.8) = 0.0359 \]
The two areas add to our desired two-tail area.
\[\begin{aligned}
P\left(|Z|>1.8\right) &= P(Z<-1.8)+P(Z>1.8) \\
&= 0.0359 + 0.0359 \\
&= 0.0718
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|>z\right) = 2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=1.8\)),
\[\begin{aligned}
P\left(|Z|<1.8\right) &= 2\cdot P(Z<-1.8) \\
&= 2\cdot0.0359 \\
&= 0.0718
\end{aligned}\]
This method is shown graphically:
Notice we need to add both tails.
Method 2: You can achieve the same result by using the following formula:
\[P\left(|Z|>z\right) = 2\cdot\left[1-P(Z<z) \right] \]
So,
\[\begin{aligned}
P\left(|Z|>1.8\right) &= 2\cdot\left[1-P(Z<1.8) \right] \\
&= 2\cdot\left[1-0.9641\right] \\
&= 2\cdot0.0359\\
&= 0.0718
\end{aligned}\]
It might be helpful to visualize with a spinner.
In a spreadsheet, you could use the NORM.DIST() function.
=2*NORM.DIST(-1.8,0,1,TRUE)
In R, you could use the pnorm function.
2*pnorm(-1.8)
Question
Determine the probability that the absolute standard normal variable is more than 0.88. In other words, evaluate \(P\left(|Z| > 0.88\right)\).
First, you need to identify that we are looking for a two-tail area (the sum of the left and right tails). This is because \(z\)-scores far from 0 satisfy \(|Z|>0.88\), while \(z\)-scores near 0 do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provides left areas, so three methods are shown to get the two-tail area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-0.88) = 0.1894\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>0.88) = 0.1894 \]
The two areas add to our desired two-tail area.
\[\begin{aligned}
P\left(|Z|>0.88\right) &= P(Z<-0.88)+P(Z>0.88) \\
&= 0.1894 + 0.1894 \\
&= 0.3788
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|>z\right) = 2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=0.88\)),
\[\begin{aligned}
P\left(|Z|<0.88\right) &= 2\cdot P(Z<-0.88) \\
&= 2\cdot0.1894 \\
&= 0.3788
\end{aligned}\]
This method is shown graphically:
Notice we need to add both tails.
Method 2: You can achieve the same result by using the following formula:
\[P\left(|Z|>z\right) = 2\cdot\left[1-P(Z<z) \right] \]
So,
\[\begin{aligned}
P\left(|Z|>0.88\right) &= 2\cdot\left[1-P(Z<0.88) \right] \\
&= 2\cdot\left[1-0.8106\right] \\
&= 2\cdot0.1894\\
&= 0.3788
\end{aligned}\]
It might be helpful to visualize with a spinner.
In a spreadsheet, you could use the NORM.DIST() function.
=2*NORM.DIST(-0.88,0,1,TRUE)
In R, you could use the pnorm function.
2*pnorm(-0.88)
Question
Determine the probability that the absolute standard normal variable is more than 0.48. In other words, evaluate \(P\left(|Z| > 0.48\right)\).
First, you need to identify that we are looking for a two-tail area (the sum of the left and right tails). This is because \(z\)-scores far from 0 satisfy \(|Z|>0.48\), while \(z\)-scores near 0 do not satisfy the inequality.
Start with a sketch.
Remember, the entire area is always 1. Most z-tables only provides left areas, so three methods are shown to get the two-tail area from a left-area (cumulative probability) table.
Method 1: First find the area of the left tail.
\[P(Z<-0.48) = 0.3156\]
Recognize normal distributions are symmetric, so we also know the area of the right tail.
\[P(Z>0.48) = 0.3156 \]
The two areas add to our desired two-tail area.
\[\begin{aligned}
P\left(|Z|>0.48\right) &= P(Z<-0.48)+P(Z>0.48) \\
&= 0.3156 + 0.3156 \\
&= 0.6312
\end{aligned}\]
This technique can be summarized with the following formula:
\[P\left(|Z|>z\right) = 2\cdot P(Z<-z)\]
assuming \(z>0\).
So, in our case (when \(z=0.48\)),
\[\begin{aligned}
P\left(|Z|<0.48\right) &= 2\cdot P(Z<-0.48) \\
&= 2\cdot0.3156 \\
&= 0.6312
\end{aligned}\]
This method is shown graphically:
Notice we need to add both tails.
Method 2: You can achieve the same result by using the following formula:
\[P\left(|Z|>z\right) = 2\cdot\left[1-P(Z<z) \right] \]
So,
\[\begin{aligned}
P\left(|Z|>0.48\right) &= 2\cdot\left[1-P(Z<0.48) \right] \\
&= 2\cdot\left[1-0.6844\right] \\
&= 2\cdot0.3156\\
&= 0.6312
\end{aligned}\]
It might be helpful to visualize with a spinner.
In a spreadsheet, you could use the NORM.DIST() function.
=2*NORM.DIST(-0.48,0,1,TRUE)
In R, you could use the pnorm function.
2*pnorm(-0.48)
Question
Determine the probability that the standard normal variable is between -1.54 and 0.09. In other words, evaluate \(P(-1.54 < Z < 0.09)\).
Determine \(z\) such that \(P(Z<z)=0.82\). In other words, what \(z\)-score is greater than \(82\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. Leftward numbers (toward \(-\infty\)) will be less than our boundary \(z\), so we shade a left region with area 0.82.
You should go to your \(z\)-table and find the \(z\)-score with the left area closest to 0.82.
\(z\)
\(P(Z<z)\)
0.89
0.8133
0.9
0.8159
0.91
0.8186
0.92
0.8212
0.93
0.8238
0.94
0.8264
It turns out the exact answer is \(z=0.9153651\), which could be found by using an inverse normal function. On a spreadsheet:
=Norm.Inv(0.82,0,1)
Using R:
qnorm(0.82)
But, the \(z\)-table is accurate enough, so I will accept either 0.91 or 0.92 (anything within 0.01 of 0.9153651).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z<z)=0.45\). In other words, what \(z\)-score is greater than \(45\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. Leftward numbers (toward \(-\infty\)) will be less than our boundary \(z\), so we shade a left region with area 0.45.
You should go to your \(z\)-table and find the \(z\)-score with the left area closest to 0.45.
\(z\)
\(P(Z<z)\)
-0.15
0.4404
-0.14
0.4443
-0.13
0.4483
-0.12
0.4522
-0.11
0.4562
-0.1
0.4602
It turns out the exact answer is \(z=-0.1256613\), which could be found by using an inverse normal function. On a spreadsheet:
=Norm.Inv(0.45,0,1)
Using R:
qnorm(0.45)
But, the \(z\)-table is accurate enough, so I will accept either -0.13 or -0.12 (anything within 0.01 of -0.1256613).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z<z)=0.52\). In other words, what \(z\)-score is greater than \(52\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. Leftward numbers (toward \(-\infty\)) will be less than our boundary \(z\), so we shade a left region with area 0.52.
You should go to your \(z\)-table and find the \(z\)-score with the left area closest to 0.52.
\(z\)
\(P(Z<z)\)
0.03
0.512
0.04
0.516
0.05
0.5199
0.06
0.5239
0.07
0.5279
0.08
0.5319
It turns out the exact answer is \(z=0.0501536\), which could be found by using an inverse normal function. On a spreadsheet:
=Norm.Inv(0.52,0,1)
Using R:
qnorm(0.52)
But, the \(z\)-table is accurate enough, so I will accept either 0.05 or 0.06 (anything within 0.01 of 0.0501536).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z<z)=0.41\). In other words, what \(z\)-score is greater than \(41\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. Leftward numbers (toward \(-\infty\)) will be less than our boundary \(z\), so we shade a left region with area 0.41.
You should go to your \(z\)-table and find the \(z\)-score with the left area closest to 0.41.
\(z\)
\(P(Z<z)\)
-0.25
0.4013
-0.24
0.4052
-0.23
0.409
-0.22
0.4129
-0.21
0.4168
-0.2
0.4207
It turns out the exact answer is \(z=-0.227545\), which could be found by using an inverse normal function. On a spreadsheet:
=Norm.Inv(0.41,0,1)
Using R:
qnorm(0.41)
But, the \(z\)-table is accurate enough, so I will accept either -0.23 or -0.22 (anything within 0.01 of -0.227545).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z<z)=0.25\). In other words, what \(z\)-score is greater than \(25\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. Leftward numbers (toward \(-\infty\)) will be less than our boundary \(z\), so we shade a left region with area 0.25.
You should go to your \(z\)-table and find the \(z\)-score with the left area closest to 0.25.
\(z\)
\(P(Z<z)\)
-0.7
0.242
-0.69
0.2451
-0.68
0.2483
-0.67
0.2514
-0.66
0.2546
-0.65
0.2578
It turns out the exact answer is \(z=-0.6744898\), which could be found by using an inverse normal function. On a spreadsheet:
=Norm.Inv(0.25,0,1)
Using R:
qnorm(0.25)
But, the \(z\)-table is accurate enough, so I will accept either -0.68 or -0.67 (anything within 0.01 of -0.6744898).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z<z)=0.75\). In other words, what \(z\)-score is greater than \(75\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. Leftward numbers (toward \(-\infty\)) will be less than our boundary \(z\), so we shade a left region with area 0.75.
You should go to your \(z\)-table and find the \(z\)-score with the left area closest to 0.75.
\(z\)
\(P(Z<z)\)
0.65
0.7422
0.66
0.7454
0.67
0.7486
0.68
0.7517
0.69
0.7549
0.7
0.758
It turns out the exact answer is \(z=0.6744898\), which could be found by using an inverse normal function. On a spreadsheet:
=Norm.Inv(0.75,0,1)
Using R:
qnorm(0.75)
But, the \(z\)-table is accurate enough, so I will accept either 0.67 or 0.68 (anything within 0.01 of 0.6744898).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z<z)=0.28\). In other words, what \(z\)-score is greater than \(28\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. Leftward numbers (toward \(-\infty\)) will be less than our boundary \(z\), so we shade a left region with area 0.28.
You should go to your \(z\)-table and find the \(z\)-score with the left area closest to 0.28.
\(z\)
\(P(Z<z)\)
-0.61
0.2709
-0.6
0.2743
-0.59
0.2776
-0.58
0.281
-0.57
0.2843
-0.56
0.2877
It turns out the exact answer is \(z=-0.5828415\), which could be found by using an inverse normal function. On a spreadsheet:
=Norm.Inv(0.28,0,1)
Using R:
qnorm(0.28)
But, the \(z\)-table is accurate enough, so I will accept either -0.59 or -0.58 (anything within 0.01 of -0.5828415).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z<z)=0.15\). In other words, what \(z\)-score is greater than \(15\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. Leftward numbers (toward \(-\infty\)) will be less than our boundary \(z\), so we shade a left region with area 0.15.
You should go to your \(z\)-table and find the \(z\)-score with the left area closest to 0.15.
\(z\)
\(P(Z<z)\)
-1.06
0.1446
-1.05
0.1469
-1.04
0.1492
-1.03
0.1515
-1.02
0.1539
-1.01
0.1562
It turns out the exact answer is \(z=-1.0364334\), which could be found by using an inverse normal function. On a spreadsheet:
=Norm.Inv(0.15,0,1)
Using R:
qnorm(0.15)
But, the \(z\)-table is accurate enough, so I will accept either -1.04 or -1.03 (anything within 0.01 of -1.0364334).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z<z)=0.32\). In other words, what \(z\)-score is greater than \(32\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. Leftward numbers (toward \(-\infty\)) will be less than our boundary \(z\), so we shade a left region with area 0.32.
You should go to your \(z\)-table and find the \(z\)-score with the left area closest to 0.32.
\(z\)
\(P(Z<z)\)
-0.49
0.3121
-0.48
0.3156
-0.47
0.3192
-0.46
0.3228
-0.45
0.3264
-0.44
0.33
It turns out the exact answer is \(z=-0.4676988\), which could be found by using an inverse normal function. On a spreadsheet:
=Norm.Inv(0.32,0,1)
Using R:
qnorm(0.32)
But, the \(z\)-table is accurate enough, so I will accept either -0.47 or -0.46 (anything within 0.01 of -0.4676988).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z<z)=0.85\). In other words, what \(z\)-score is greater than \(85\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. Leftward numbers (toward \(-\infty\)) will be less than our boundary \(z\), so we shade a left region with area 0.85.
You should go to your \(z\)-table and find the \(z\)-score with the left area closest to 0.85.
\(z\)
\(P(Z<z)\)
1.01
0.8438
1.02
0.8461
1.03
0.8485
1.04
0.8508
1.05
0.8531
1.06
0.8554
It turns out the exact answer is \(z=1.0364334\), which could be found by using an inverse normal function. On a spreadsheet:
=Norm.Inv(0.85,0,1)
Using R:
qnorm(0.85)
But, the \(z\)-table is accurate enough, so I will accept either 1.03 or 1.04 (anything within 0.01 of 1.0364334).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z>z)=0.36\). In other words, what \(z\)-score is less than \(36\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
But, the \(z\)-table is accurate enough, so I will accept either 0.35 or 0.36 (anything within 0.01 of 0.3584588).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z>z)=0.9\). In other words, what \(z\)-score is less than \(90\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
But, the \(z\)-table is accurate enough, so I will accept either -1.29 or -1.28 (anything within 0.01 of -1.2815516).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z>z)=0.73\). In other words, what \(z\)-score is less than \(73\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
But, the \(z\)-table is accurate enough, so I will accept either -0.62 or -0.61 (anything within 0.01 of -0.612813).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z>z)=0.89\). In other words, what \(z\)-score is less than \(89\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
But, the \(z\)-table is accurate enough, so I will accept either -1.23 or -1.22 (anything within 0.01 of -1.2265281).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z>z)=0.9\). In other words, what \(z\)-score is less than \(90\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
But, the \(z\)-table is accurate enough, so I will accept either -1.29 or -1.28 (anything within 0.01 of -1.2815516).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z>z)=0.32\). In other words, what \(z\)-score is less than \(32\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
But, the \(z\)-table is accurate enough, so I will accept either 0.46 or 0.47 (anything within 0.01 of 0.4676988).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z>z)=0.79\). In other words, what \(z\)-score is less than \(79\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
But, the \(z\)-table is accurate enough, so I will accept either -0.81 or -0.8 (anything within 0.01 of -0.8064212).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z>z)=0.76\). In other words, what \(z\)-score is less than \(76\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
But, the \(z\)-table is accurate enough, so I will accept either -0.71 or -0.7 (anything within 0.01 of -0.7063026).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z>z)=0.19\). In other words, what \(z\)-score is less than \(19\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
But, the \(z\)-table is accurate enough, so I will accept either 0.87 or 0.88 (anything within 0.01 of 0.8778963).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(Z>z)=0.46\). In other words, what \(z\)-score is less than \(46\)% of standard normal values? (Answers within 0.01 from the correct value will be marked correct.)
But, the \(z\)-table is accurate enough, so I will accept either 0.1 or 0.11 (anything within 0.01 of 0.1004337).
You might find it helpful to visualize with a spinner.
Question
Determine \(z\) such that \(P(|Z|<z)=0.38\). In other words, how far from 0 should boundaries be set such that 38% of standard normal values are between those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Determine \(z\) such that \(P(|Z|<z)=0.38\). In other words, how far from 0 should boundaries be set such that 38% of standard normal values are between those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Determine \(z\) such that \(P(|Z|<z)=0.96\). In other words, how far from 0 should boundaries be set such that 96% of standard normal values are between those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Determine \(z\) such that \(P(|Z|<z)=0.54\). In other words, how far from 0 should boundaries be set such that 54% of standard normal values are between those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Determine \(z\) such that \(P(|Z|<z)=0.24\). In other words, how far from 0 should boundaries be set such that 24% of standard normal values are between those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Determine \(z\) such that \(P(|Z|<z)=0.36\). In other words, how far from 0 should boundaries be set such that 36% of standard normal values are between those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Determine \(z\) such that \(P(|Z|<z)=0.96\). In other words, how far from 0 should boundaries be set such that 96% of standard normal values are between those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Determine \(z\) such that \(P(|Z|<z)=0.9\). In other words, how far from 0 should boundaries be set such that 90% of standard normal values are between those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Determine \(z\) such that \(P(|Z|<z)=0.62\). In other words, how far from 0 should boundaries be set such that 62% of standard normal values are between those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Determine \(z\) such that \(P(|Z|<z)=0.76\). In other words, how far from 0 should boundaries be set such that 76% of standard normal values are between those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Determine \(z\) such that \(P(|Z|>z)=0.06\). In other words, how far from 0 should boundaries be set such that 6% of standard normal values are outside those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. The total two-tail area is 0.06, so each tail has half that area.
Method 1: Determine the area of each tail and the center. Both tails have the same area, and all three areas add to 1. Thus,
\[P(Z>z) = \frac{0.06}{2} = 0.03\]
Determine \(z\) such that \(P(|Z|>z)=0.38\). In other words, how far from 0 should boundaries be set such that 38% of standard normal values are outside those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. The total two-tail area is 0.38, so each tail has half that area.
Method 1: Determine the area of each tail and the center. Both tails have the same area, and all three areas add to 1. Thus,
\[P(Z>z) = \frac{0.38}{2} = 0.19\]
Determine \(z\) such that \(P(|Z|>z)=0.5\). In other words, how far from 0 should boundaries be set such that 50% of standard normal values are outside those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. The total two-tail area is 0.5, so each tail has half that area.
Method 1: Determine the area of each tail and the center. Both tails have the same area, and all three areas add to 1. Thus,
\[P(Z>z) = \frac{0.5}{2} = 0.25\]
Determine \(z\) such that \(P(|Z|>z)=0.58\). In other words, how far from 0 should boundaries be set such that 58% of standard normal values are outside those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. The total two-tail area is 0.58, so each tail has half that area.
Method 1: Determine the area of each tail and the center. Both tails have the same area, and all three areas add to 1. Thus,
\[P(Z>z) = \frac{0.58}{2} = 0.29\]
Determine \(z\) such that \(P(|Z|>z)=0.42\). In other words, how far from 0 should boundaries be set such that 42% of standard normal values are outside those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. The total two-tail area is 0.42, so each tail has half that area.
Method 1: Determine the area of each tail and the center. Both tails have the same area, and all three areas add to 1. Thus,
\[P(Z>z) = \frac{0.42}{2} = 0.21\]
Determine \(z\) such that \(P(|Z|>z)=0.4\). In other words, how far from 0 should boundaries be set such that 40% of standard normal values are outside those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. The total two-tail area is 0.4, so each tail has half that area.
Method 1: Determine the area of each tail and the center. Both tails have the same area, and all three areas add to 1. Thus,
\[P(Z>z) = \frac{0.4}{2} = 0.2\]
Determine \(z\) such that \(P(|Z|>z)=0.66\). In other words, how far from 0 should boundaries be set such that 66% of standard normal values are outside those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. The total two-tail area is 0.66, so each tail has half that area.
Method 1: Determine the area of each tail and the center. Both tails have the same area, and all three areas add to 1. Thus,
\[P(Z>z) = \frac{0.66}{2} = 0.33\]
Determine \(z\) such that \(P(|Z|>z)=0.74\). In other words, how far from 0 should boundaries be set such that 74% of standard normal values are outside those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. The total two-tail area is 0.74, so each tail has half that area.
Method 1: Determine the area of each tail and the center. Both tails have the same area, and all three areas add to 1. Thus,
\[P(Z>z) = \frac{0.74}{2} = 0.37\]
Determine \(z\) such that \(P(|Z|>z)=0.14\). In other words, how far from 0 should boundaries be set such that 14% of standard normal values are outside those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. The total two-tail area is 0.14, so each tail has half that area.
Method 1: Determine the area of each tail and the center. Both tails have the same area, and all three areas add to 1. Thus,
\[P(Z>z) = \frac{0.14}{2} = 0.07\]
Determine \(z\) such that \(P(|Z|>z)=0.1\). In other words, how far from 0 should boundaries be set such that 10% of standard normal values are outside those boundaries? (Answers within 0.01 from the correct value will be marked correct.)
Start with a sketch. The total two-tail area is 0.1, so each tail has half that area.
Method 1: Determine the area of each tail and the center. Both tails have the same area, and all three areas add to 1. Thus,
\[P(Z>z) = \frac{0.1}{2} = 0.05\]
A scientist is investigating whether a chemical may effect the growth of an organism. Under the control conditions (no chemical), the organism grows to a mean mass of \(\mu_0 = 14\) grams with a standard deviation of \(\sigma_0 =3.9\) grams. These population parameters are known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=37\) times. In that sample, the masses have a mean \(\bar{x}=12.68\).
The scientist wonders if this sample mean is significantly different from \(\mu_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample mean as far (or farther) from \(\mu_0\) due to chance alone.
It is common to compare the \(p\)-value to 0.05.
If \(p\text{-value} < 0.05\) then the difference of means is “significant” and warrants publication.
If \(p\text{-value} \ge 0.05\) then “not significant”. It is possible a larger sample size will allow detection of a significant difference, or maybe the chemical has no effect.
The difference is significant; the chemical seems to alter the growth of the organism.
Question
A scientist is investigating whether a chemical may effect the growth of an organism. Under the control conditions (no chemical), the organism grows to a mean mass of \(\mu_0 = 63.6\) grams with a standard deviation of \(\sigma_0 =19\) grams. These population parameters are known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=79\) times. In that sample, the masses have a mean \(\bar{x}=58.43\).
The scientist wonders if this sample mean is significantly different from \(\mu_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample mean as far (or farther) from \(\mu_0\) due to chance alone.
It is common to compare the \(p\)-value to 0.05.
If \(p\text{-value} < 0.05\) then the difference of means is “significant” and warrants publication.
If \(p\text{-value} \ge 0.05\) then “not significant”. It is possible a larger sample size will allow detection of a significant difference, or maybe the chemical has no effect.
The difference is significant; the chemical seems to alter the growth of the organism.
Question
A scientist is investigating whether a chemical may effect the growth of an organism. Under the control conditions (no chemical), the organism grows to a mean mass of \(\mu_0 = 16.5\) grams with a standard deviation of \(\sigma_0 =5.4\) grams. These population parameters are known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=88\) times. In that sample, the masses have a mean \(\bar{x}=15.14\).
The scientist wonders if this sample mean is significantly different from \(\mu_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample mean as far (or farther) from \(\mu_0\) due to chance alone.
It is common to compare the \(p\)-value to 0.05.
If \(p\text{-value} < 0.05\) then the difference of means is “significant” and warrants publication.
If \(p\text{-value} \ge 0.05\) then “not significant”. It is possible a larger sample size will allow detection of a significant difference, or maybe the chemical has no effect.
The difference is significant; the chemical seems to alter the growth of the organism.
Question
A scientist is investigating whether a chemical may effect the growth of an organism. Under the control conditions (no chemical), the organism grows to a mean mass of \(\mu_0 = 31.7\) grams with a standard deviation of \(\sigma_0 =8.2\) grams. These population parameters are known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=30\) times. In that sample, the masses have a mean \(\bar{x}=28.39\).
The scientist wonders if this sample mean is significantly different from \(\mu_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample mean as far (or farther) from \(\mu_0\) due to chance alone.
It is common to compare the \(p\)-value to 0.05.
If \(p\text{-value} < 0.05\) then the difference of means is “significant” and warrants publication.
If \(p\text{-value} \ge 0.05\) then “not significant”. It is possible a larger sample size will allow detection of a significant difference, or maybe the chemical has no effect.
The difference is significant; the chemical seems to alter the growth of the organism.
Question
A scientist is investigating whether a chemical may effect the growth of an organism. Under the control conditions (no chemical), the organism grows to a mean mass of \(\mu_0 = 79.4\) grams with a standard deviation of \(\sigma_0 =22.5\) grams. These population parameters are known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=71\) times. In that sample, the masses have a mean \(\bar{x}=85.22\).
The scientist wonders if this sample mean is significantly different from \(\mu_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample mean as far (or farther) from \(\mu_0\) due to chance alone.
It is common to compare the \(p\)-value to 0.05.
If \(p\text{-value} < 0.05\) then the difference of means is “significant” and warrants publication.
If \(p\text{-value} \ge 0.05\) then “not significant”. It is possible a larger sample size will allow detection of a significant difference, or maybe the chemical has no effect.
The difference is significant; the chemical seems to alter the growth of the organism.
Question
A scientist is investigating whether a chemical may effect the growth of an organism. Under the control conditions (no chemical), the organism grows to a mean mass of \(\mu_0 = 59.3\) grams with a standard deviation of \(\sigma_0 =10.5\) grams. These population parameters are known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=65\) times. In that sample, the masses have a mean \(\bar{x}=56.97\).
The scientist wonders if this sample mean is significantly different from \(\mu_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample mean as far (or farther) from \(\mu_0\) due to chance alone.
It is common to compare the \(p\)-value to 0.05.
If \(p\text{-value} < 0.05\) then the difference of means is “significant” and warrants publication.
If \(p\text{-value} \ge 0.05\) then “not significant”. It is possible a larger sample size will allow detection of a significant difference, or maybe the chemical has no effect.
The difference is not significant; we don’t know whether the chemical alters the growth of the organism.
Question
A scientist is investigating whether a chemical may effect the growth of an organism. Under the control conditions (no chemical), the organism grows to a mean mass of \(\mu_0 = 87.9\) grams with a standard deviation of \(\sigma_0 =15.9\) grams. These population parameters are known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=67\) times. In that sample, the masses have a mean \(\bar{x}=84.46\).
The scientist wonders if this sample mean is significantly different from \(\mu_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample mean as far (or farther) from \(\mu_0\) due to chance alone.
It is common to compare the \(p\)-value to 0.05.
If \(p\text{-value} < 0.05\) then the difference of means is “significant” and warrants publication.
If \(p\text{-value} \ge 0.05\) then “not significant”. It is possible a larger sample size will allow detection of a significant difference, or maybe the chemical has no effect.
The difference is not significant; we don’t know whether the chemical alters the growth of the organism.
Question
A scientist is investigating whether a chemical may effect the growth of an organism. Under the control conditions (no chemical), the organism grows to a mean mass of \(\mu_0 = 13.3\) grams with a standard deviation of \(\sigma_0 =3\) grams. These population parameters are known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=88\) times. In that sample, the masses have a mean \(\bar{x}=13.85\).
The scientist wonders if this sample mean is significantly different from \(\mu_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample mean as far (or farther) from \(\mu_0\) due to chance alone.
It is common to compare the \(p\)-value to 0.05.
If \(p\text{-value} < 0.05\) then the difference of means is “significant” and warrants publication.
If \(p\text{-value} \ge 0.05\) then “not significant”. It is possible a larger sample size will allow detection of a significant difference, or maybe the chemical has no effect.
The difference is not significant; we don’t know whether the chemical alters the growth of the organism.
Question
A scientist is investigating whether a chemical may effect the growth of an organism. Under the control conditions (no chemical), the organism grows to a mean mass of \(\mu_0 = 36.2\) grams with a standard deviation of \(\sigma_0 =9.8\) grams. These population parameters are known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=74\) times. In that sample, the masses have a mean \(\bar{x}=34.16\).
The scientist wonders if this sample mean is significantly different from \(\mu_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample mean as far (or farther) from \(\mu_0\) due to chance alone.
It is common to compare the \(p\)-value to 0.05.
If \(p\text{-value} < 0.05\) then the difference of means is “significant” and warrants publication.
If \(p\text{-value} \ge 0.05\) then “not significant”. It is possible a larger sample size will allow detection of a significant difference, or maybe the chemical has no effect.
The difference is not significant; we don’t know whether the chemical alters the growth of the organism.
Question
A scientist is investigating whether a chemical may effect the growth of an organism. Under the control conditions (no chemical), the organism grows to a mean mass of \(\mu_0 = 52.8\) grams with a standard deviation of \(\sigma_0 =14.9\) grams. These population parameters are known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=51\) times. In that sample, the masses have a mean \(\bar{x}=56.45\).
The scientist wonders if this sample mean is significantly different from \(\mu_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample mean as far (or farther) from \(\mu_0\) due to chance alone.
It is common to compare the \(p\)-value to 0.05.
If \(p\text{-value} < 0.05\) then the difference of means is “significant” and warrants publication.
If \(p\text{-value} \ge 0.05\) then “not significant”. It is possible a larger sample size will allow detection of a significant difference, or maybe the chemical has no effect.
The difference is not significant; we don’t know whether the chemical alters the growth of the organism.
Question
A scientist is investigating whether a chemical may effect the survival rate of an organism. Under the control conditions (no chemical), the organism has a survival rate of \(p_0=0.9571\). This value is known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=70\) times. In that sample, the survival rate is \(\hat{p} = 0.9\).
The scientist wonders if this survival rate is significantly different from \(p_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample proportion as far (or farther) from \(p_0\) due to chance alone.
\[p\text{-value} ~=~ P\left(\big|Z\big| > \frac{\big|\hat{p}-p_0\big|}{\sqrt{\frac{p_0(1-p_0)}{n}}} \right) \]
The difference is significant; the chemical seems to alter the growth of the organism.
0.0183914
Question
A scientist is investigating whether a chemical may effect the survival rate of an organism. Under the control conditions (no chemical), the organism has a survival rate of \(p_0=0.0714\). This value is known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=42\) times. In that sample, the survival rate is \(\hat{p} = -0.0238\).
The scientist wonders if this survival rate is significantly different from \(p_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample proportion as far (or farther) from \(p_0\) due to chance alone.
\[p\text{-value} ~=~ P\left(\big|Z\big| > \frac{\big|\hat{p}-p_0\big|}{\sqrt{\frac{p_0(1-p_0)}{n}}} \right) \]
The difference is significant; the chemical seems to alter the growth of the organism.
0.0165723
Question
A scientist is investigating whether a chemical may effect the survival rate of an organism. Under the control conditions (no chemical), the organism has a survival rate of \(p_0=0.7778\). This value is known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=90\) times. In that sample, the survival rate is \(\hat{p} = 0.8667\).
The scientist wonders if this survival rate is significantly different from \(p_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample proportion as far (or farther) from \(p_0\) due to chance alone.
\[p\text{-value} ~=~ P\left(\big|Z\big| > \frac{\big|\hat{p}-p_0\big|}{\sqrt{\frac{p_0(1-p_0)}{n}}} \right) \]
The difference is significant; the chemical seems to alter the growth of the organism.
0.0424892
Question
A scientist is investigating whether a chemical may effect the survival rate of an organism. Under the control conditions (no chemical), the organism has a survival rate of \(p_0=0.4045\). This value is known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=89\) times. In that sample, the survival rate is \(\hat{p} = 0.2921\).
The scientist wonders if this survival rate is significantly different from \(p_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample proportion as far (or farther) from \(p_0\) due to chance alone.
\[p\text{-value} ~=~ P\left(\big|Z\big| > \frac{\big|\hat{p}-p_0\big|}{\sqrt{\frac{p_0(1-p_0)}{n}}} \right) \]
The difference is significant; the chemical seems to alter the growth of the organism.
0.0307313
Question
A scientist is investigating whether a chemical may effect the survival rate of an organism. Under the control conditions (no chemical), the organism has a survival rate of \(p_0=0.4667\). This value is known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=45\) times. In that sample, the survival rate is \(\hat{p} = 0.2889\).
The scientist wonders if this survival rate is significantly different from \(p_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample proportion as far (or farther) from \(p_0\) due to chance alone.
\[p\text{-value} ~=~ P\left(\big|Z\big| > \frac{\big|\hat{p}-p_0\big|}{\sqrt{\frac{p_0(1-p_0)}{n}}} \right) \]
The difference is significant; the chemical seems to alter the growth of the organism.
0.0168142
Question
A scientist is investigating whether a chemical may effect the survival rate of an organism. Under the control conditions (no chemical), the organism has a survival rate of \(p_0=0.0448\). This value is known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=67\) times. In that sample, the survival rate is \(\hat{p} = 0.0896\).
The scientist wonders if this survival rate is significantly different from \(p_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample proportion as far (or farther) from \(p_0\) due to chance alone.
\[p\text{-value} ~=~ P\left(\big|Z\big| > \frac{\big|\hat{p}-p_0\big|}{\sqrt{\frac{p_0(1-p_0)}{n}}} \right) \]
The difference is not significant; we don’t know whether the chemical alters the growth of the organism.
0.0762825
Question
A scientist is investigating whether a chemical may effect the survival rate of an organism. Under the control conditions (no chemical), the organism has a survival rate of \(p_0=0.75\). This value is known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=24\) times. In that sample, the survival rate is \(\hat{p} = 0.9583\).
The scientist wonders if this survival rate is significantly different from \(p_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample proportion as far (or farther) from \(p_0\) due to chance alone.
\[p\text{-value} ~=~ P\left(\big|Z\big| > \frac{\big|\hat{p}-p_0\big|}{\sqrt{\frac{p_0(1-p_0)}{n}}} \right) \]
The difference is significant; the chemical seems to alter the growth of the organism.
0.0184408
Question
A scientist is investigating whether a chemical may effect the survival rate of an organism. Under the control conditions (no chemical), the organism has a survival rate of \(p_0=0.7831\). This value is known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=83\) times. In that sample, the survival rate is \(\hat{p} = 0.6988\).
The scientist wonders if this survival rate is significantly different from \(p_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample proportion as far (or farther) from \(p_0\) due to chance alone.
\[p\text{-value} ~=~ P\left(\big|Z\big| > \frac{\big|\hat{p}-p_0\big|}{\sqrt{\frac{p_0(1-p_0)}{n}}} \right) \]
The difference is not significant; we don’t know whether the chemical alters the growth of the organism.
0.0623927
Question
A scientist is investigating whether a chemical may effect the survival rate of an organism. Under the control conditions (no chemical), the organism has a survival rate of \(p_0=0.4483\). This value is known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=29\) times. In that sample, the survival rate is \(\hat{p} = 0.2759\).
The scientist wonders if this survival rate is significantly different from \(p_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample proportion as far (or farther) from \(p_0\) due to chance alone.
\[p\text{-value} ~=~ P\left(\big|Z\big| > \frac{\big|\hat{p}-p_0\big|}{\sqrt{\frac{p_0(1-p_0)}{n}}} \right) \]
The difference is not significant; we don’t know whether the chemical alters the growth of the organism.
0.061928
Question
A scientist is investigating whether a chemical may effect the survival rate of an organism. Under the control conditions (no chemical), the organism has a survival rate of \(p_0=0.8202\). This value is known precisely because the organism has been grown under control conditions many many times.
The scientist has only grown the organism under experimental conditions (with chemical) \(n=89\) times. In that sample, the survival rate is \(\hat{p} = 0.7416\).
The scientist wonders if this survival rate is significantly different from \(p_0\). To investigate this, the scientist will determine the \(p\)-value. The \(p\)-value represents the probability of getting a sample proportion as far (or farther) from \(p_0\) due to chance alone.
\[p\text{-value} ~=~ P\left(\big|Z\big| > \frac{\big|\hat{p}-p_0\big|}{\sqrt{\frac{p_0(1-p_0)}{n}}} \right) \]
The difference is not significant; we don’t know whether the chemical alters the growth of the organism.
0.0534937
Question
A doctor runs a controlled experiment. The participants are randomly assigned to two groups: control and treatment. The participants in the control group are given a placebo. The participants in the treatment group are given a drug. After several months, each participant’s triglyceride level is measured (in mg/dL).
x1
x2
553.7
576.6
577.8
633.6
573.0
589.7
564.3
565.6
516.6
553.3
536.2
531.3
556.3
607.6
525.2
620.1
507.2
521.9
601.2
585.6
557.1
667.7
582.9
578.0
573.3
498.8
519.6
641.4
510.4
635.0
575.8
549.0
550.7
555.8
585.0
596.3
561.3
575.3
567.5
560.8
529.5
486.7
584.9
533.2
489.7
652.9
You are asked to perform a two-tail two-sample Welch’s \(t\) test to determine whether there is a significant difference of means in the two samples.
Determine the \(p\)-value. Answers within a tolerance of 0.005 will be accepted.
Solution
Use a spreadsheet. Copy the data over to a new sheet.
You just need to use T.TEST with the proper settings.
Question
A doctor runs a controlled experiment. The participants are randomly assigned to two groups: control and treatment. The participants in the control group are given a placebo. The participants in the treatment group are given a drug. After several months, each participant’s triglyceride level is measured (in mg/dL).
x1
x2
476.3
543.7
589.6
539.6
660.5
551.1
583.9
539.1
658.2
555.8
658.9
528.4
651.2
567.3
663.5
561.9
555.8
605.0
513.7
582.6
558.8
550.0
655.6
593.0
594.1
617.4
565.8
592.4
498.6
551.1
555.9
575.8
558.7
572.0
699.7
538.8
569.8
529.1
585.4
569.7
677.3
553.9
594.7
593.4
629.1
572.3
You are asked to perform a two-tail two-sample Welch’s \(t\) test to determine whether there is a significant difference of means in the two samples.
Determine the \(p\)-value. Answers within a tolerance of 0.005 will be accepted.
Solution
Use a spreadsheet. Copy the data over to a new sheet.
You just need to use T.TEST with the proper settings.
Question
A doctor runs a controlled experiment. The participants are randomly assigned to two groups: control and treatment. The participants in the control group are given a placebo. The participants in the treatment group are given a drug. After several months, each participant’s triglyceride level is measured (in mg/dL).
x1
x2
571.3
409.4
489.4
470.7
475.8
543.6
537.5
565.3
380.0
625.8
558.0
569.2
553.0
479.9
503.0
506.3
501.9
612.4
548.4
613.5
555.8
505.5
501.3
503.8
498.8
566.0
533.0
538.0
437.3
529.7
523.9
517.4
560.2
467.4
498.9
498.6
452.3
471.0
You are asked to perform a two-tail two-sample Welch’s \(t\) test to determine whether there is a significant difference of means in the two samples.
Determine the \(p\)-value. Answers within a tolerance of 0.005 will be accepted.
Solution
Use a spreadsheet. Copy the data over to a new sheet.
You just need to use T.TEST with the proper settings.
Question
A doctor runs a controlled experiment. The participants are randomly assigned to two groups: control and treatment. The participants in the control group are given a placebo. The participants in the treatment group are given a drug. After several months, each participant’s triglyceride level is measured (in mg/dL).
x1
x2
385.1
396.8
408.1
400.9
365.6
437.4
362.8
407.9
373.5
403.1
404.5
397.3
392.6
412.0
387.0
378.2
378.4
392.1
398.8
409.6
387.0
405.7
391.3
384.0
400.5
382.4
378.4
381.2
406.8
413.5
378.3
428.8
371.6
368.3
394.2
425.9
408.3
402.0
400.5
384.6
You are asked to perform a two-tail two-sample Welch’s \(t\) test to determine whether there is a significant difference of means in the two samples.
Determine the \(p\)-value. Answers within a tolerance of 0.005 will be accepted.
Solution
Use a spreadsheet. Copy the data over to a new sheet.
You just need to use T.TEST with the proper settings.
Question
A doctor runs a controlled experiment. The participants are randomly assigned to two groups: control and treatment. The participants in the control group are given a placebo. The participants in the treatment group are given a drug. After several months, each participant’s triglyceride level is measured (in mg/dL).
x1
x2
405.7
435.4
302.6
355.8
340.7
308.0
353.1
384.4
393.1
369.5
331.3
322.1
298.7
311.8
376.3
316.8
348.2
345.4
346.5
296.4
456.2
252.8
316.8
325.7
401.4
327.1
391.7
307.9
372.6
323.6
380.9
281.0
297.6
329.7
397.8
385.0
320.6
364.6
318.9
330.2
351.7
358.2
324.0
You are asked to perform a two-tail two-sample Welch’s \(t\) test to determine whether there is a significant difference of means in the two samples.
Determine the \(p\)-value. Answers within a tolerance of 0.005 will be accepted.
Solution
Use a spreadsheet. Copy the data over to a new sheet.
You just need to use T.TEST with the proper settings.
Question
A doctor runs a controlled experiment. The participants are randomly assigned to two groups: control and treatment. The participants in the control group are given a placebo. The participants in the treatment group are given a drug. After several months, each participant’s triglyceride level is measured (in mg/dL).
x1
x2
562.3
578.4
567.2
403.3
652.4
500.2
590.3
439.9
512.4
504.0
487.3
663.7
591.6
433.2
518.4
497.6
525.3
488.3
590.1
550.9
505.8
532.7
512.2
487.9
568.4
394.4
501.6
553.5
462.7
531.6
564.4
492.9
514.5
520.0
493.3
492.9
514.5
607.8
587.3
601.0
602.7
636.2
579.5
You are asked to perform a two-tail two-sample Welch’s \(t\) test to determine whether there is a significant difference of means in the two samples.
Determine the \(p\)-value. Answers within a tolerance of 0.005 will be accepted.
Solution
Use a spreadsheet. Copy the data over to a new sheet.
You just need to use T.TEST with the proper settings.
Question
A doctor runs a controlled experiment. The participants are randomly assigned to two groups: control and treatment. The participants in the control group are given a placebo. The participants in the treatment group are given a drug. After several months, each participant’s triglyceride level is measured (in mg/dL).
x1
x2
323.4
305.6
233.0
305.4
302.9
278.7
253.3
267.5
317.1
332.4
267.4
282.8
270.4
276.2
258.0
314.0
290.5
284.7
262.1
273.5
290.1
307.0
282.3
249.1
280.8
308.9
236.4
291.8
265.6
313.5
319.0
305.6
311.4
226.1
289.4
290.3
320.1
262.0
276.3
267.6
You are asked to perform a two-tail two-sample Welch’s \(t\) test to determine whether there is a significant difference of means in the two samples.
Determine the \(p\)-value. Answers within a tolerance of 0.005 will be accepted.
Solution
Use a spreadsheet. Copy the data over to a new sheet.
You just need to use T.TEST with the proper settings.
Question
A doctor runs a controlled experiment. The participants are randomly assigned to two groups: control and treatment. The participants in the control group are given a placebo. The participants in the treatment group are given a drug. After several months, each participant’s triglyceride level is measured (in mg/dL).
x1
x2
345.1
363.5
304.4
340.1
341.9
371.6
338.0
317.8
321.2
353.9
346.0
337.5
369.3
368.5
319.8
358.1
332.5
305.6
313.8
344.8
347.3
385.0
345.9
384.1
324.2
349.7
324.5
362.1
361.5
348.2
330.1
354.7
374.5
404.6
380.4
392.3
278.7
362.7
You are asked to perform a two-tail two-sample Welch’s \(t\) test to determine whether there is a significant difference of means in the two samples.
Determine the \(p\)-value. Answers within a tolerance of 0.005 will be accepted.
Solution
Use a spreadsheet. Copy the data over to a new sheet.
You just need to use T.TEST with the proper settings.
Question
A doctor runs a controlled experiment. The participants are randomly assigned to two groups: control and treatment. The participants in the control group are given a placebo. The participants in the treatment group are given a drug. After several months, each participant’s triglyceride level is measured (in mg/dL).
x1
x2
258.4
257.1
212.2
289.8
214.7
275.6
278.8
246.6
248.1
275.5
265.7
317.0
227.2
255.6
234.3
242.4
214.7
197.3
256.7
193.9
211.5
258.8
278.6
282.7
223.6
218.1
208.5
353.5
241.6
323.0
197.6
279.7
221.2
246.3
217.2
308.1
242.1
232.9
232.6
286.4
210.6
229.8
270.2
181.8
236.9
318.8
251.9
You are asked to perform a two-tail two-sample Welch’s \(t\) test to determine whether there is a significant difference of means in the two samples.
Determine the \(p\)-value. Answers within a tolerance of 0.005 will be accepted.
Solution
Use a spreadsheet. Copy the data over to a new sheet.
You just need to use T.TEST with the proper settings.
Question
A doctor runs a controlled experiment. The participants are randomly assigned to two groups: control and treatment. The participants in the control group are given a placebo. The participants in the treatment group are given a drug. After several months, each participant’s triglyceride level is measured (in mg/dL).
x1
x2
340.6
200.7
320.2
316.9
201.7
396.0
288.1
230.9
228.8
202.8
321.2
361.1
265.2
260.9
189.8
251.0
266.8
300.2
268.5
328.7
255.1
273.7
275.9
370.6
297.2
253.5
277.4
254.6
214.2
376.8
256.2
374.8
287.0
235.1
187.9
235.6
179.5
249.8
267.3
You are asked to perform a two-tail two-sample Welch’s \(t\) test to determine whether there is a significant difference of means in the two samples.
Determine the \(p\)-value. Answers within a tolerance of 0.005 will be accepted.
Solution
Use a spreadsheet. Copy the data over to a new sheet.
You just need to use T.TEST with the proper settings.
